The general solution of $sin 2 x + sin 4 x + sin 6 x = 0$ is/are

n π \pm \frac{π}{6}, n \in Z

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\frac{n π}{4}, n \in Z

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2 n π \pm \frac{2 π}{3}, n \in Z

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n π \pm \frac{π}{3}, n \in Z

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Solution

The correct option is D $n π \pm \frac{π}{3}, n \in Z$
Given: $sin 2 x + sin 4 x + sin 6 x = 0$
$\Rightarrow (sin 6 x + sin 2 x) + sin 4 x = 0 \Rightarrow 2 sin 4 x cos 2 x + sin 4 x = 0 \Rightarrow sin 4 x (2 cos 2 x + 1) = 0 \Rightarrow sin 4 x = 0 or 2 cos 2 x + 1 = 0 \Rightarrow sin 4 x = 0 or cos 2 x = - \frac{1}{2} \Rightarrow sin 4 x = 0 or cos 2 x = cos \frac{2 π}{3} \Rightarrow 4 x = n π or 2 x = (2 n π \pm \frac{2 π}{3}), n \in Z$

Hence, the general solution is
$x = \frac{n π}{4} or x = (n π \pm \frac{π}{3}), n \in Z$

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